Let $f:\mathbb R \to \mathbb R$ be a continuous function such that for all real $x$ , lim$_{n\to \infty} f(nx)=0$ , then how to prove that lim$_{x \to \infty}f(x)=0$ ?

Please help and please don't use Baire Category theorem , I don't know it

  • $\begingroup$ A change of variables should do it (in a way that I think is more clear than using the same variable $x$ in both cases). Here is a sketch: consider the limit $Lim_{y\rightarrow \infty} f(y)=0$. For any y, for any fixed non-zero $x$ ,you can choose $n$, so that $nx>y$. Then, if $f(nx) \rightarrow 0$ , same goes for $f(y)$. More specifically, if $f(nx) \rightarrow 0$ , this means there is an integer $N$ so that $|f(nx)|< \epsilon $ whenever $n>N$. Now, for given $y$ choose $n$ large-enough so that $nx>y$. $\endgroup$
    – user99680
    Jun 3 '14 at 4:45
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    $\begingroup$ @user99680: the problem I see with your suggestion is that $nx>y$ and $|f(nx)|<\epsilon$ does not imply that $|f(y)|<\epsilon$. $\endgroup$ Jun 3 '14 at 5:59
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    $\begingroup$ I have an idea for a proof that would need the following lemma: given any disjoint set of open intervals $(a_j,b_j)$ where the $a_j$ tend to infinity, there exists a real number $x$ such that $\{x,2x,3x,\dots\}$ has infinitely many points in common with the union of the $(a_j,b_j)$. Can anyone prove (or disprove) this lemma? $\endgroup$ Jun 3 '14 at 6:05
  • $\begingroup$ it's stupid of me (and there really may by a short and simple proof), but unless you're forbidden to use Baire theorem (thanks for the hint :), you can learn it - its proof is simple. Then you'd just consider the closed sets $S_n=\{x\in\mathbb R;(\forall k\geq n)\,|f(kx)|\leq\epsilon\}$ ($S_1\subset S_2\subset S_3\subset\dots$), by your assumption $\bigcup_n S_n=\mathbb R$, by Baire one of $S_n$'s contains an open interval $(a,b)$ , $a,b>0$, and from there you show that for $N$ big enough, $|f(x)|\leq \epsilon$ $\forall x>N$. $\endgroup$
    – user8268
    Jun 3 '14 at 7:15
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    $\begingroup$ The keyword Croft's Lemma will also be helpful! $\endgroup$ Jun 3 '14 at 8:38

Here, I will prove Greg Martin's lemma, and thereby proving the theorem.

Suppose that the conclusion $f(x)\rightarrow 0$ is not true, then there is an $\epsilon>0$ such that $|f(x_j)|>\epsilon$ for infinitely many $x_j$ where $x_j\rightarrow\infty$ as $j\rightarrow\infty$.

By continuity of $f$, we have a disjoint set of open intervals $(a_j, b_j)$ such that $$ |f(x)| > \epsilon$$ for all $x$ in the union of $(a_j, b_j)$, and $a_j\rightarrow \infty$ as $j\rightarrow\infty$.

Without loss of generality, we can assume that $a_1>0$.

Let $[a,b]\subset (a_1, b_1)$ with $a\neq b$.

Now, here's the KEY IDEA:

There exist some $n$ such that $[na,nb]$ intersects $(a_j,b_j)$ for some $j>1$. This is possible because $n(b-a)\rightarrow \infty$ as $n\rightarrow\infty$. Let $n_1$ be the first $n$ with this property, and let $j_1$ be the corresponding $j$ for intersecting $(a_{j },b_{j })$.

The intersection must be nontrivial interval of positive length, therefore we can find a nontrivial closed interval $[c_1,d_1]$ such that $$[c_1,d_1]\subset [a,b]\cap (a_{j_1}/n_1, b_{j_1}/n_1).$$

The next step is almost identical to the previous step, but we require $j>j_1$. Indeed, there exist some $n$ such that $[nc_1,nd_1]$ intersects $(a_j,b_j)$ for some $j>j_1$. Then it gives a nontrivial closed interval $[c_2, d_2]$ such that $$[c_2, d_2]\subset [c_1, d_1]\cap (a_{j_2}/n_2, b_{j_2}/n_2).$$

By this construction, we obtain a nested intervals $$ [a,b]\supset [c_1,d_1] \supset [c_2,d_2]\supset \cdots $$

Then let $c\in \bigcap_j [c_j, d_j]$. For this $c$, we must have $|f(n_jc)|> \epsilon$ for all $j$. This contradicts the assumption $f(nc)\rightarrow 0$ as $n\rightarrow\infty$.

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    $\begingroup$ Nice job! Rather than "this is possible because $n(b-a)\to\infty$ as $n\to\infty$", which doesn't (as far as I can see) force one of the intervals to intersect a possibly sparse set, I think it's more salient to say "this is possible because as soon as $n>1/(b-a)$, then the intervals $[na,nb]$, $[(n+1)a,(n+1)b]$, ... all overlap and hence cover the entire ray $[na,\infty)$". $\endgroup$ Jun 3 '14 at 17:26

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